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Are the following 2 equations lorentz invariant?

  1. Aug 13, 2013 #1
    [itex]\partial[/itex]μ[itex]\phi[/itex][itex]\partial[/itex]μ[itex]\phi[/itex]

    and

    [itex]\partial[/itex]μ[itex]\partial[/itex]μ[itex]\phi[/itex]

    with [itex]\phi[/itex](x) a scalar field
     
  2. jcsd
  3. Aug 13, 2013 #2

    Mentz114

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    Suppose ##\partial_{\mu'}=\Lambda^\mu_{\mu'}\partial_{\mu}## and ##\partial^{\mu'}=\Lambda^{\mu'}_{\mu}\partial^{\mu}##.
    Then
    ##\partial_{\mu'}\partial^{\mu'}\phi=\Lambda^\mu_{\mu'}\partial_{\mu} \Lambda^{\mu'}_{\mu}\partial^{\mu}\phi=\Lambda^\mu_{\mu'} \Lambda^{\mu'}_{\mu}\partial_{\mu}\partial^{\mu}\phi##.

    So the scalar is invariant if ##\Lambda^\mu_{\mu'}## is the inverse of ## \Lambda^{\mu'}_{\mu}##.

    Now, ##\Lambda^\mu_{\mu'}## is not a tensor but a transformation matrix. It so happens that the inverse of the LT is found by reversing the sign of the boost velocity. This operation ##\Lambda_b^{b'}=\eta_{ba}\Lambda_{a'}^a\eta^{a'b'}## also reverses the sign. So for the Lorentz transformation ##\Lambda^\mu_{\mu'}## is the inverse of ## \Lambda^{\mu'}_{\mu}##.
     
  4. Aug 13, 2013 #3
    So the answer is yes?!
    I actually thought the second one was lorentzinvariant because:

    [itex]\partial[/itex]μ[itex]\partial[/itex]μ is the d'Alembertian which I thought was Lorentzinvariant, while a scalar field is always Lorentzinvariant?

    And what about the other equation? I'm not sure what the influence of the second scalar field in the equation yields.
     
  5. Aug 13, 2013 #4

    Mentz114

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    The answer is 'yes' for both equations. If you write out the first expression in terms of the transformed coords, there is another cancellation of ##\Lambda^\mu_{\mu'}\Lambda_\mu^{\mu'}##.

    Rank-0 tensors ( ie no indexes) are always invariant. Differentiating a rank-0 tensor creates a new tensor with rank-1. Scalars formed by contractions of tensors are invariant.
     
  6. Aug 13, 2013 #5
    Thank you very much! :)
     
  7. Aug 13, 2013 #6

    vanhees71

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    One should, however, be a bit more careful with the indices in the Lorentz transformations and write them also with proper order. Let
    [tex]{\Lambda^{\mu}}_{\nu}[/tex]
    the Lorentz-transformation matrix for contravariant vector/tensor components, i.e.,
    [tex]x'^{\mu}={\Lambda^{\mu}}_{\nu} x^{\nu}.[/tex]
    A Lorentz transform obeys
    [tex]\eta_{\mu \nu} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\eta_{\rho \sigma},[/tex]
    or
    [tex]\eta_{\mu \nu} \eta^{\rho \alpha} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.[/tex]
    This implies
    [tex]\eta^{\rho \alpha} \eta_{\mu \nu} {L^{\mu}}_{\rho}={(L^{-1})^{\alpha}}_{\nu}={L_{\mu}}^{\alpha}.[/tex]
     
  8. Aug 13, 2013 #7

    Mentz114

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    Thanks for clarifying that.
     
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